# Ex 9.4, 1 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 10, 2019 by Teachoo

Last updated at Dec. 10, 2019 by Teachoo

Transcript

Ex 9.4, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : ππ¦/ππ₯=(1 β cosβ‘π₯)/(1 + cosβ‘π₯ ) ππ¦/ππ₯ = (1 β cosβ‘π₯)/(1 + cosβ‘π₯ ) We know that cos 2x = 2cos2 x β 1 Putting x = π₯/2 cos 2π₯/2 = 2 cos2 π₯/2 β 1 cos x = 2 cos2 π₯/2 β 1 1 + cos x = 2cos2 π₯/2 We know cos 2x = 1 β 2sin2 x Putting x = π₯/2 cos2 2π₯/2 = 1 β 2 sin2 π₯/2 cos x = 1 β 2sin2 π₯/2 1 β cos x = 2sin2 π₯/2 Putting values in equation ππ¦/ππ₯ = (2 γsin^2 γβ‘γπ₯/2γ )/(2 γcos^2 γβ‘γπ₯/2γ ) ππ¦ = (γsin^2 γβ‘γπ₯/2γ )/cos^(2 )β‘γπ₯/2γ ππ₯ ππ¦ = tan2 π₯/2 ππ₯ Putting tan2 π₯/2 = sec2 π₯/2 β 1 ππ¦ = ("sec2 " π₯/2 " β 1" )ππ₯ Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x β 1 Putting x = π₯/2 tan2 π₯/2 = sec2 π₯/2 β 1 β«1βππ¦ = β«1β(π ππ2 π₯/2β1) ππ₯ y = β«1βγπ ππ2 π₯/2 ππ₯β β«1βππ₯γ y = 1/(1/2) tan π₯/2 β x + C y = 2 tan π₯/2 β x + c y = 2 tan π₯/2 β x + C y = π tan π/π β x + C is the general solution (As β«1βγπ ππ2 ππ₯ππ₯= 1/πγ tan ax + C)

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Chapter 9 Class 12 Differential Equations

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.